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Linear independence of basis vectors

  1. Nov 14, 2005 #1
    How do I prove the linear independence of the standard basis vectors? My book is helpful by giving the definition of linear independence and a couple examples, but never once shows how to prove that they are linearly independent.
    I know that the list of standard basis vectors is linearly independent if:
    The only choice of a_1, a_2, ... a_m that makes a_1v_1+a_2v_2+...+a_mv_m=0 is a_1=a_2=...=a_m=0.
    But i don't know where to go from there .. any help would be appreciated :confused:
    Last edited: Nov 14, 2005
  2. jcsd
  3. Nov 14, 2005 #2
    Use contradiction. If they were linearly dependent, then by your definition, at least one of the vectors could be written as a linear combination of others. Is this true of the standard basis ?
  4. Nov 14, 2005 #3


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    For example, in [itex]\mathbb{R}^3 [/itex], the standard basis is:

    [tex]\left\{ {\left( {1,0,0} \right),\left( {0,1,0} \right),\left( {0,0,1} \right)} \right\}[/tex]

    This basis is linearly independant if, as you say:

    [tex]a\left( {1,0,0} \right) + b\left( {0,1,0} \right) + c\left( {0,0,1} \right) = \left( {0,0,0} \right)[/tex]

    implies that [itex]a = b = c = 0[/itex].

    Well, solve the condition for a, b and c and see if you can find anything else besides the solution we expect.
    Can you now see how it will be for [itex]\mathbb{R}^n [/itex] in general?
  5. Nov 14, 2005 #4
    Thank you so much .. it actually makes sense now .. Something about being out of school with a broken pelvis means that it's harder to understand what they do in class without you ... thanks!
  6. Nov 14, 2005 #5


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    No problem :smile:
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